Question: $h(n) = -5n^{2}+f(n)$ $f(t) = 3t$ $g(n) = -6n^{3}+2n^{2}-f(n)$ $ g(h(1)) = {?} $
Solution: First, let's solve for the value of the inner function, $h(1)$ . Then we'll know what to plug into the outer function. $h(1) = -5(1^{2})+f(1)$ To solve for the value of $h$ , we need to solve for the value of $f(1)$ $f(1) = (3)(1)$ $f(1) = 3$ That means $h(1) = -5(1^{2})+3$ $h(1) = -2$ Now we know that $h(1) = -2$ . Let's solve for $g(h(1))$ , which is $g(-2)$ $g(-2) = -6(-2)^{3}+2(-2)^{2}-f(-2)$ To solve for the value of $g$ , we need to solve for the value of $f(-2)$ $f(-2) = (3)(-2)$ $f(-2) = -6$ That means $g(-2) = -6(-2)^{3}+2(-2)^{2}-(-6)$ $g(-2) = 62$